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If we consider the simplest model of helium atom, namely infinitely massive nucleus with two non-interacting electrons, we'll get an additive combination of two spectra of a hydrogen-like atom (with $Z=2$). The wavefunctions will be also just (anti)symmetrized products of hydrogen-like single-electron wavefunctions. All is well: up until double ionization there exist doubly-excited states, all of which are truly bound, i.e. square integrable (by construction) — despite being above single ionization threshold.

Now if we include electron-electron repulsion, things get complicated. From multiple papers$^\dagger$ discussing doubly-excited states in helium I learned that such states are actually not square-integrable, and are usually calculated by complex rotation method. The resonances thus obtained appear to have nonzero imaginary part of energy, which, as I understand, implies non-square-integrability of actual states of non-rotated Hamiltonian.

My question is now: what is actual reason for these states to stop being square-integrable — is it mere presence of electron-electron repulsion (so that repulsion potential could be made e.g. $\propto\operatorname{sech}(r_{12})$, and the non-integrability would be preserved)? Does this situation require potential energy to be unbounded in some locus of configuration space, or can we reproduce this situation with a smooth total potential of the system?


$^\dagger$Examples of such papers:

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As you rightly note, all of the doubly excited states of helium are above the single-ionization threshold, and this means that they are auto-ionizing: they have enough energy to slingshot one of the electrons into the continuum and, given half the chance, will proceed to do so.

The only reason that this doesn't happen in the naive theory is that the electrons simply don't 'know' about the existence of the other guy to an extent that's sufficient for the transfer of actual quantum mechanical energy between them. That is, you can add self-consistent mean-value-of-the-density contributions to the (now nonlinear and nonlocal) hamiltonian for each electron, but you are still, by the simple fact that you're solving single-particle Schrödinger equations, explicitly forbidding the kinds of configuration interactions that make autoionizing states spring to life.

As such, any sort of entangling interaction between the two electrons will act as a link between your (formerly) bound states and the continuum, and it will force the eigenstates to be a linear combination of both: that is, the bound state becomes a non-square-integrable resonance embedded in the continuum, with an imaginary part of the eigenvalue which requires the state to decay over time as population goes away towards infinity.

The example you mentioned, $\mathrm{sech}(|\mathbf r_1-\mathbf r_2|)$, is a good candidate, as would be a softened Coulomb potential, but you could also use e.g. a contact interaction, $\delta(\mathbf r_1-\mathbf r_2)$, and get much the same results. (In fact, those would be pretty interesting to see, but I'm not aware of any references and it sounds like a hard thing to search for.)

If you want the abstract theory, there's probably no place better than the original Fano paper on Fano resonances,

U. Fano. Effects of configuration interaction on intensities and phase shifts. Phys. Rev. 124, 1866 (1961), eprints.

It has aged a bit, but it is still pretty readable, and the depth of its arguments makes up for its slightly outdated notation. In particular, it is very general regarding what sorts of particle-particle coupling would give rise to the autoionizing character of those states: all you need is a nonzero entangling matrix element, and it will happen. (Even better, so long as that matrix element is reasonably flat along the relevant bit of continuum, you also get the Fano line profile as a universal characteristic, with only the $q$ parameter tuning the shape.)

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  • $\begingroup$ From your explanation I get that even if we soften all the singularities in the potential, the imaginary parts of resonance energies will remain nonzero, is it right? $\endgroup$ – Ruslan Feb 22 '17 at 12:54
  • $\begingroup$ Yes, that is correct. All you need for autoionization is enough energy, a suitable continuum at the same (Hartree-Fock) energy, and an entangling operator between those two sectors. $\endgroup$ – Emilio Pisanty Feb 22 '17 at 12:55

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